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Trac #33100: use a more reliable abs() replacement in is_hermitian().
Some finite fields for which is_hermitian() otherwise works are missing abs(). Here we use the square root of z*z.conjugate() as a replacement, which appears to work in all of the cases we care about. Note that this would probably do the wrong thing if you tried to use a nonzero tolerance with an (exact) finite field.
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